11.4: Solving Equations of the Form ax = b and x/a = b
- be familiar with the multiplication/division property of equality
- be able to solve equations of the form \(ax = b\) and \(\dfrac{x}{a} = b\)
- be able to use combined techniques to solve equations
Multiplication/ Division Property of Equality
Recall that the equal sign of an equation indicates that the number represented by the expression on the left side is the same as the number represented by the expression on the right side. From this, we can suggest the multiplication/division property of equality.
Multiplication/Division Property of
Equality
Given any equation,
-
We can obtain an equivalent equation by
multiplying both sides
of the equation by the
same nonzero
number, that is, if \(c \ne 0\). then \(a = b\) is equivalent to
\(a \cdot c = b \cdot c\) -
We can obtain an equivalent equation by
dividing both sides
of the equation by the
same nonzero number
, that is, if \(c \ne 0\), then \(a = b\) is equivalent to
\(\dfrac{a}{c} = \dfrac{b}{c}\)
The multiplication/division property of equality can be used to undo an association with a number that multiplies or divides the variable.
Use the multiplication / division property of equality to solve each equation.
\(6y = 54\)
Solution
6 is associated with y by multiplication. Undo the association by dividing both sides by 6
\(\begin{array} {rcl} {\dfrac{6y}{6}} & = & {\dfrac{54}{6}} \\ {\dfrac{\cancel{6} y}{\cancel{6}}} & = & {\dfrac{\begin{array} {c} {^9} \\ {\cancel{54}} \end{array}}{\cancel{6}}} \\ {y} & = & {9} \end{array}\)
Check: When \(y = 9\)
\(6y = 54\)
becomes
a true statement.
The solution to \(6y = 54\) is \(y = 9\).
\(\dfrac{x}{-2} = 27\).
Solution
-2 is associated with \(x\) by division. Undo the association by multiplying both sides by -2.
\((-2) \dfrac{x}{-2} = (-2) 27\)
\((\cancel{-2}) \dfrac{x}{\cancel{-2}} = (-2) 27\)
\(x = -54\)
Check: When \(x = -54\).
\(\dfrac{x}{-2} = 27\)
becomes
a true statement.
The solution to \(\dfrac{x}{-2} = 27\) is \(x = -54\)
\(\dfrac{3a}{7} = 6\).
Solution
We will examine two methods for solving equations such as this one.
Method 1: Use of dividing out common factors.
\(\dfrac{3a}{7} = 6\)
7 is associated with \(a\) by division. Undo the association by multiplying both sides by 7.
\(7 \cdot \dfrac{3a}{7} = 7 \cdot 6\)
Divide out the 7's
\(\cancel{7} \cdot \dfrac{3a}{\cancel{7}} = 42\)
\(3a = 42\)
3 is associated with \(a\) by multiplication. Undo the association by dviding both sides by 3.
\(\dfrac{3a}{3} = \dfrac{42}{3}\)
\(\dfrac{\cancel{3} a}{\cancel{3}} = 14\)
\(a = 14\)
Check: When \(a = 14\).
\(\dfrac{3a}{7} = 6\)
becomes
a true statement.
The solution to \(\dfrac{3a}{7} = 6\) is \(a = 14\).
Method 2: Use of reciprocals
Recall that if the product of two numbers is 1, the numbers are reciprocals . Thus \(\dfrac{3}{7}\) and \(\dfrac{7}{3}\) are reciprocals.
\(\dfrac{3a}{7} = 6\)
Multiply both sides of the equation by \(\dfrac{7}{3}\), the reciprocal of \(\dfrac{3}{7}\).
\(\dfrac{7}{3} \cdot \dfrac{3a}{7} = \dfrac{7}{3} \cdot 6\)
\(\dfrac{\begin{array} {c} {^1} \\ {\cancel{7}} \end{array}}{\begin{array} {c} {\cancel{3}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^1} \\ {\cancel{3} a} \end{array}}{\begin{array} {c} {\cancel{7}} \\ {^1} \end{array}} = \dfrac{7}{\begin{array} {c} {\cancel{3}} \\ {^1} \end{array}} \cdot \dfrac{\begin{array} {c} {^2} \\ {\cancel{6}} \end{array}}{1}\)
\(1 \cdot a = 14\)
\(a = 14\)
Notice that we get the same solution using either method.
\(-8x = 24\).
Solution
-8 is associated with \(x\) by multiplication. Undo the association by dividing both sides by -8.
\(\dfrac{-8x}{-8} = \dfrac{24}{-8}\)
\(x = -3\)
Check: When \(x = -3\).
\(-8x = 24\)
becomes
a true statement.
\(-x = 7\).
Solution
Since \(-x\) is actually \(-1 \cdot x\) and \((-1)(-1) = 1\). We can isolate \(x\) by multiplying both sides of the equation by -1.
\((-1)(-x) = -1 \cdot 7\).
\(x = -7\)
Check: When \(x= 7\).
\(-x = 7\)
becomes
The solution to \(-x = 7\) is \(x = -7\).
Practice Set A
Use the multiplication/division property of equality to solve each equation. Be sure to check each solution.
\(7x = 21\)
- Answer
-
\(x = 3\)
Practice Set A
\(-5x = 65\)
- Answer
-
\(x = -13\)
Practice Set A
\(\dfrac{x}{4} = -8\)
- Answer
-
\(x = -32\)
Practice Set A
\(\dfrac{3x}{8} = 6\)
- Answer
-
\(x = 16\)
Practice Set A
\(-y = 3\)
- Answer
-
\(y = -3\)
Practice Set A
\(-k = -2\)
- Answer
-
\(k = 2\)
Combining Techniques in Equation Solving
Having examined solving equations using the addition/subtraction and the multiplication/division principles of equality, we can combine these techniques to solve more complicated equations.
When beginning to solve an equation such as \(6x - 4 = -16\), it is helpful to know which property of equality to use first, addition/subtraction or multiplication/division. Recalling that in equation solving we are trying to isolate the variable (disassociate numbers from it), it is helpful to note the following.
To associate numbers and letters, we use the order of operations.
- Multiply/divide
- Add/subtract
To undo an association between numbers and letters, we use the order of operations in reverse.
- Add/subtract
- Multiply/divide
Solve each equation. (In these example problems, we will not show the checks.)
\(6x - 4 = -16\)
Solution
-4 is associated with xx by subtraction. Undo the association by adding 4 to both sides.
\(6x - 4 + 4 = -16 + 4\)
\(6x = -12\)
6 is associated with \(x\) by multiplication. Undo the association by dividing both sides by 6
\(\dfrac{6x}{6} = \dfrac{-12}{6}\)
\(x = -2\)
\(-8k + 3 = -45\)
Solution
3 is associated with \(k\) by addition. Undo the association by subtracting 3 from both sides.
\(-8k + 3 - 3 = -45 - 3\)
\(-8k = -48\)
-8 is associated with \(k\) by multiplication. Undo the association by dividing both sides by -8.
\(\dfrac{-8k}{-8} = \dfrac{-48}{-8}\)
\(k = 6\)
\(5m - 6 - 4m = 4m - 8 + 3m\).
Solution
Begin by solving this equation by combining like terms.
\(m - 6 = 7m - 8\) Choose a side on which to isolate
m
. Since 7 is greater than 1, we'll isolate
m
on the right side.
Subtract
m
from
both
sides.
\(-m - 6 - m = 7m - 8 - m\)
\(-6 = 6m - 8\)
8 is associated with m by subtraction. Undo the association by adding 8 to both sides.
\(-6 + 8 = 6m - 8 + 8\)
\(2 = 6m\)
6 is associated with m by multiplication. Undo the association by dividing both sides by 6.
\(\dfrac{2}{6} = \dfrac{6m}{6}\) Reduce,
\(\dfrac{1}{3} = m\)
Notice that if we had chosen to isolate m on the left side of the equation rather than the right side, we would have proceeded as follows:
\(m - 6 = 7m - 8\)
Subtract \(7m\) from both sides.
\(m - 6 - 7m = 7m - 8 - 7m\)
\(-6m - 6 = -8\)
Add 6 to both sides,
\(-6m - 6 + 6 = -8 + 6\)
\(-6m = -2\)
Divide both sides by -6.
\(\dfrac{-6m}{-6} = \dfrac{-2}{-6}\)
\(m = \dfrac{1}{3}\)
This is the same result as with the previous approach.
\(\dfrac{8x}{7} = -2\)
Solution
7 is associated with \(x\) by division. Undo the association by multiplying both sides by 7.
\(\cancel{7} \cdot \dfrac{8x}{\cancel{7}} = 7(-2)\)
\(7 \cdot \dfrac{8x}{7} = -14\)
\(8x = -14\)
8 is associated with \(x\) by multiplication. Undo the association by dividing both sides by 8.
\(\dfrac{\cancel{8} x}{\cancel{8}} = \dfrac{-7}{4}\)
\(x = \dfrac{-7}{4}\)
Practice Set B
Solve each equation. Be sure to check each solution.
\(5m + 7 = -13\)
- Answer
-
\(m = -4\)
Practice Set B
\(-3a - 6 = 9\)
- Answer
-
\(a = -5\)
Practice Set B
\(2a + 10 - 3a = 9\)
- Answer
-
\(a = 1\)
Practice Set B
\(11x - 4 - 13x = 4x + 14\)
- Answer
-
\(x = -3\)
Practice Set B
\(-3m + 8 = -5m + 1\)
- Answer
-
\(m = -\dfrac{7}{2}\)
Practice Set B
\(5y + 8y - 11 = -11\)
- Answer
-
\(y = 0\)
Exercises
Solve each equation. Be sure to check each result.
Exercise \(\PageIndex{1}\)
\(7x = 42\)
- Answer
-
\(x = 6\)
Exercise \(\PageIndex{2}\)
\(8x = 81\)
Exercise \(\PageIndex{3}\)
\(10 x = 120\)
- Answer
-
\(x = 12\)
Exercise \(\PageIndex{4}\)
\(11 x = 121\)
Exercise \(\PageIndex{5}\)
\(-6a = 48\)
- Answer
-
\(a = -8\)
Exercise \(\PageIndex{6}\)
\(-9y = 54\)
Exercise \(\PageIndex{7}\)
\(-3y = -42\)
- Answer
-
\(y = 14\)
Exercise \(\PageIndex{8}\)
\(-5a = -105\)
Exercise \(\PageIndex{9}\)
\(2m = -62\)
- Answer
-
\(m = -31\)
Exercise \(\PageIndex{10}\)
\(3m = -54\)
Exercise \(\PageIndex{11}\)
\(\dfrac{x}{4} = 7\)
- Answer
-
\(x = 28\)
Exercise \(\PageIndex{12}\)
\(\dfrac{y}{3} = 11\)
Exercise \(\PageIndex{13}\)
\(\dfrac{-z}{6} = -14\)
- Answer
-
\(z = 84\)
Exercise \(\PageIndex{14}\)
\(\dfrac{-w}{5} = 1\)
Exercise \(\PageIndex{1}\)
\(3m - 1 = -13\)
- Answer
-
\(m = -4\)
Exercise \(\PageIndex{15}\)
\(4x + 7 = -17\)
Exercise \(\PageIndex{1}\)
\(2 + 9x = -7\)
- Answer
-
\(x = -1\)
Exercise \(\PageIndex{16}\)
\(5 - 11x = 27\)
Exercise \(\PageIndex{17}\)
\(32 = 4y + 6\)
- Answer
-
\(y = \dfrac{13}{2}\)
Exercise \(\PageIndex{18}\)
\(-5 + 4 = -8m + 1\)
Exercise \(\PageIndex{19}\)
\(3k + 6 = 5k + 10\)
- Answer
-
\(k = -2\)
Exercise \(\PageIndex{20}\)
\(4a + 16 = 6a + 8a + 6\)
Exercise \(\PageIndex{21}\)
\(6x + 5 + 2x - 1 = 9x - 3x + 15\)
- Answer
-
\(x = \dfrac{11}{2}\) or \(5 \dfrac{1}{2}\)
Exercise \(\PageIndex{22}\)
\(-9y - 8 + 3y + 7 = -7y + 8y - 5y + 9\)
Exercise \(\PageIndex{23}\)
\(-3a = a + 5\)
- Answer
-
\(a = -\dfrac{5}{4})
Exercise \(\PageIndex{24}\)
\(5b = -2b + 8b + 1\)
Exercise \(\PageIndex{25}\)
\(-3m + 2 - 8m - 4 = -14m + m - 4\)
- Answer
-
\(m = -1\)
Exercise \(\PageIndex{26}\)
\(5a + 3 = 3\)
Exercise \(\PageIndex{27}\)
\(7x + 3x = 0\)
- Answer
-
\(x = 0\)
Exercise \(\PageIndex{28}\)
\(7g + 4 - 11g = -4g + 1 + g\)
Exercise \(\PageIndex{29}\)
\(\dfrac{5a}{7} = 10\)
- Answer
-
\(a = 14\)
Exercise \(\PageIndex{30}\)
\(\dfrac{2m}{9} = 4\)
Exercise \(\PageIndex{31}\)
\(\dfrac{3x}{4} = \dfrac{9}{2}\)
- Answer
-
\(x = 6\)
Exercise \(\PageIndex{32}\)
\(\dfrac{8k}{3} = 32\)
Exercise \(\PageIndex{33}\)
\(\dfrac{3a}{8} - \dfrac{3}{2} = 0\)
- Answer
-
\(a = 4\)
Exercise \(\PageIndex{34}\)
\(\dfrac{5m}{6} - \dfrac{25}{3} = 0\)
Exercises for Review
Exercise \(\PageIndex{35}\)
Use the distributive property to compute \(40 \cdot 28\)
- Answer
-
\(40(30 - 2) = 1200 - 80 = 1120\)
Exercise \(\PageIndex{36}\)
Approximating \(\pi\) by 3.14, find the approximate circumference of the circle.
Exercise \(\PageIndex{37}\)
Find the area of the parallelogram.
- Answer
-
220 sq cm
Exercise \(\PageIndex{38}\)
Find the value of \(\dfrac{-3(4 - 15) - 2}{-5}\)
Exercise \(\PageIndex{39}\)
Solve the equation \(x - 14 + 8 = -2\).
- Answer
-
\(x = 4\)